We have now talked about most of the aspects of a typical solar system installation, but there is one aspect we haven’t touched upon yet. Determining cable sizes is another very critical thing that you have to work into the design of you system to not only make sure that the efficiency of your system is optimal, but to also make sure that you don’t run any risk of cables melting or burning because they are too thin for the job. The converse is also true – you don’t necessarily want to “over-spec” the cables that you acquire too much; copper cabling is pretty expensive these days and you want to make sure that you buy just the right cables for the specific application, including off course a little bit of a buffer for safety and possible future expansion.

*Things to consider when determining cable sizes*

There are really four important things you have to be concerned about when thinking about getting the correct cables to connect everything in your system. They are:

- Current carrying capacity of the cable
- The average ambient temperature that your cable will be working in
- Voltage drop over the length of the cable
- The load factor of the specific load that you are going to use the cable for

The first thing, and what I consider to be the most important thing, is the ** current carrying capacity of the cable**. This is a number that is specified in ampere (A). You should always stay well within the limits of the current carrying capacity of the cable, no matter which part of your system it is connecting. This also goes for the cables you use to do earth connections!

OK, to see if the cable has enough current carrying capacity, you simply have to determine what the voltage at the feed or source side of the cable is, what the approximate length and ambient temperature of the cable will be, and then also what the load is on the other side. Let’s use an example of a 230 volt supply, and let’s say that you have a 1 kilowatt load that you want to drive on the other end of a 35 meter cable, operating at 35 degrees Celsius (86 degrees Fahrenheit).

From the basic principles that we discussed before, you will remember that Ohm’s law says that P = I x V, (there are other aspects/variations of Ohm’s law as well but don’t fret over that now!) which simply means that the power (P, or wattage) is equal to the current (I, measured in amps) times the voltage (V, measured in volts). We have the value for the voltage, which is 230 volts, and we also have the size of the load, which in the case of our example is 1 kilowatt, or 1000 watt. From the equation above, we can then figure out the current, I = P / V, (trust me, the formula is manipulated to reflect this with a little basic math!) or rather, the current (I, in amps) that the load will draw through the cable is equal to the power (P = 1000 watt) divided by the voltage (V = 230 volt). Just hold on tight here, it’s actually really simple – that means that I (the current) is equal to 1000/230 = 4.35 amps. See, really not that difficult at all!

In this example, it means that we have to provide a cable to connect this load of 1000 watt to the supply of 230 volt, which will be able to carry at least 4.35 amps, over the 35 meter length of the cable. It is good practice to add about 20% additional capacity to the cable size. To do the math, you can either add 20% to the load (20% of 1000 watt is 200 watt, added to the load gives a total of 1200 watt. I = 1200 / 230 = 5.22 amps), OR you can just add it to the minimum current carrying capacity of the cable that you choose (4.35 amps x 1.2 for the additional 20% = 5.22 amps). Including a 20% “buffer” or safety factor, means that we have to choose a cable that has a minimum current carrying capacity of 5.22 amps. [Not finished yet, we still have to do a few other checks regarding the cable size, which I discuss below!]

From Table 1 below, we now have to choose the cable that is closest to the current carrying capacity that we need. The cable with the specification closest to what we need is the first one, with current carrying capacity of 11 amp, and size (thickness of the copper conductors) of 1 square millimeter (the British measures are also indicated in the table for convenience, in this case it is 0.0015 square inches per conductor, also with a current carrying capacity of 11 amps).

** Ambient temperature** also plays a role. (This is the second factor you need to consider). Table 2 below gives the “temperature factor” for a range of temperatures. Choose the average temperature in your area that is closest to the next highest temperature setting listed in the table. If you look at Table 1, you will notice that all ratings are given for the cable at 30 degrees Celsius (86 degrees Fahrenheit). To compensate for the cable operating at a temperature of 35 degrees Celsius, not 30 degrees, we have to multiply the current carrying capacity of the cable that we chose, by the

**. In this case, from Table 2, the temperature factor is 0.97. If we multiply the current carrying capacity (11 amp) by 0.97, we get 10.6 amp. Because our requirement is only 5.22 amp, and the temperature compensated current carrying capacity of the cable is 10.6 amp, our cable is clearly still good enough according to the math up to here.**

*Temperature Factor*The third thing we have to check to make sure that our cable will be fine for this application, is to make sure that the ** voltage drop** over the length of the cable is also acceptable. The accepted norm is that the maximum allowable voltage drop from the connection point at the source of electricity, to the furthest end of the cable where you connect the load should not exceed 2.5%

*(Just for interest sake: This proposed limit for voltage drop over cable lengths comes from the IEEE rule B-23, which is a set of standards for just about everything electrical, managed by the IEEE – Institute for Electrical & Electronics Engineers. The maximum allowable voltage drop for a main feeder cable from the “meter box/point of supply” to the “distribution box” is 1.25%).*In our example we are working with a 230 volt system supply, and we assume that the load will be the furthest end of the system, which means that the maximum allowable voltage drop will be 2.5% of 230 volt, 2.5% x 230 = 5.75 volt. So, 5.75 volt is the maximum allowable voltage drop for this specific cable in this specific application.

From Table 3 below, we can see that the voltage drop for our chosen cable is 41 millivolt per amp-meter. (Table 4 is given for your convenience if you live in a part of the world where the imperial system is used for measurement – The exact same principles apply for all calculations). This voltage drop of 41 millivolt per meter, is given as a standard for 1 ampere of current flowing through the cable over a 1 meter length. Now, from the calculation we did above, you can see that we will not be drawing only 1 ampere through the cable, but actually, we will be drawing 5.22 amps through the cable (including off course our “safety factor” of 20%). We therefore need to multiply the 41 millivolt by 5.22 to figure out the real voltage drop we will get per meter, with our load of 1000 watt connected at the end of the cable. Hope this makes it clear! With the specification in the British/Imperial system, this calculation works exactly the same way. The numbers differ slightly between the metric system and the imperial system because the cable thicknesses are slightly different.

So, based on the above, to calculate the voltage drop, in the case of our example above, over the length of the cable (35 meters), we need to multiply the specified voltage drop in millivolt by the maximum current carrying capacity of the cable and by the length of the cable.

That means that the total voltage drop in our chosen cable over the 35 meter length will be (41/1000) x 11 x 35 = 15.8 volt. Now, before you rush off to go and buy your cable, there is the last factor that you have to take into account. This is the ** load factor** – it is a measure of the current that we are actually going to draw through the cable, in relation to the maximum current carrying capacity of the cable that we chose. So, let’s now do the calculation for the load factor as well.

The ** load factor** is simply the current we expect to draw through the cable (in the case of our example this value is 5.22 amps), divided by the

__maximum temperature adjusted current carrying capacity__of the cable. The load factor will therefore be 5.22/10.6 = 0.49. We now need to multiply the voltage drop by the load factor to determine the ACTUAL voltage drop that we will get in this specific installation. This calculation is: Voltage drop (actual) = calculated maximum voltage drop (15.8 volt) x Load factor (0.49) = 7.74 volt. You can clearly see that this ACTUAL voltage drop from our calculation is more than the maximum allowable voltage drop specified earlier on (5.75 volt). This means that we have to choose the next “thicker” cable and do the calculations again to make sure that this time it will all work out.

Because our new choice of cable size will be thicker, our current carrying capacity will be higher, and we therefore know that our previous calculations of the minimum size of the required cable according to the ** current carrying capacity** will still be valid. We do still have to re-do the temperature compensation calculation, as well as the voltage drop calculations, and the load factor calculations to make sure that the next size cable will be suitable for the application. The voltage drop in this next “thicker” size cable, according to Table 3 above is 28 millivolt, at a current rating of 13 amp [This is the data from the second line in the Table, 13 amps, 28 millivolts, and cable size of 1.5 square millimeters]. The temperature compensated maximum current carrying capacity of this new cable will be 0.97 x 13 amp = 12.6 amp. Our calculation will then be: Voltage drop = (28/1000) x 12.6 x 35 = 12.35 volt. Now we need to bring the load factor into play again. Our

**calculation will be ACTUAL current (5.22 amp) / maximum current carrying capacity (12.6 amp) = 0.41. The calculation will then be Voltage drop (ACTUAL) = 12.7 volt x Load factor (0.41) = 5.21 volt. The maximum allowable voltage drop that we discussed before is 5.75 volt, which means that our cable is now within all allowable requirements – the cable is “thick” enough to handle the load, and to also deal with the voltage drop, as well as the elevated ambient temperature. The cable you have to use for this application is therefore a 1.5 square millimeter cable, with the specifications indicated in Table 3 above. The cables specified in the tables are all pretty standard right across the world from a sizing perspective. If all else fails guys, rather buy a slightly thicker cable than what your calculations tell you to buy – that is always the safer route to follow.**

*new load factor*Even to do the calculations for an entire electrical system, you simply follow the same principles to calculate the “size” of the cable required for each branch of each circuit. When you do the design of cable sizes for internal wiring in a home, factory or other building, you also have to be aware of the minimum specifications that are laid down by local or national authorities in the country or local area where you live. These regulations may indicate “thicker” cable for such internal wiring applications, which is essentially set up so that there is always a higher “safety factor” in these systems to keep all of us safe.

In a future post I am planning to put a whole system together and do all the necessary calculations to show you how the whole lot is put together.

*Simplified summary*

This is a very short summary of how to determine the size or gauge of the cables that you will need to connect everything in your solar system:

- Determine the source voltage
- Determine the size of the load at the end of the cable
- Measure the length of the cable and figure out the average ambient temperature at which this cable will be used
- From the table get the relevant data for your cable that you chose
- Do the calculations as discussed above
- You are now ready to go out and buy the most appropriate size of cable that you will need for the installation of your solar system
- Enjoy and have fun
- Always remember, if you have any doubts, questions or concerns o CONTACT A PROFESSIONAL IN YOUR AREA AND ASK FOR ADVICE BEFORE YOU DO ANYTHING! (This will always be my little “mantra” for everyone – please, always take care, make use of the safety instructions given in a previous post, and always, always check and ask for a professional’s input if you have any doubts)

That’s it for this week…

Once again, if anything is not clear to you, or if you have any specific questions regarding the discussion this week, please reply to this email and I will gladly assist you to figure it out. Until next week, when we will have a look at an entire system and all the requirements and issues involved. I may have to split that discussion over more than one post to make sure you don’t get bored in the process. We will take it as it comes. Happy calculating!